Rewrite the function by completing the square. $g(x)= 4 x^{2} -28 x +49$ $g(x)=$
Solution: $\begin{aligned} g(x)&=4 x^2 -28 x +49 \\\\ &=4 \left(x^2 -7 x\right) +49 \end{aligned}$ Now we want to complete $x^2 -7 x$ into a perfect square. To do that, we should add $\left(\dfrac{{-7}}{2}\right)^2={\dfrac{49}{4}}$ to it: $x^2{-7}x+{\dfrac{49}{4}}=\left(x -\dfrac{7}{2}\right)^2$ We add ${\dfrac{49}{4}}$ inside the parentheses, and subtract ${4}\cdot{\dfrac{49}{4}}$ outside them, to keep the expression equivalent. $\begin{aligned} &\phantom{=}{4} \left(x^2 -7 x\right) +49 \\\\ &={4}\left(x^2 -7 x+{\dfrac{49}{4}}\right) +49 -{4}\cdot{\dfrac{49}{4}} \\\\ &=4 \left(x -\dfrac{7}{2}\right)^2 +49 -49 \\\\ &=4 \left(x -\dfrac{7}{2}\right)^2 +0 \end{aligned}$ In conclusion, the function after completing the square is written as: $g(x)=4 \left(x -\dfrac{7}{2}\right)^2 +0$